Set Theory Problem: equality of functions

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SUMMARY

The discussion centers on proving the equality of functions F and G, defined as F = G if and only if domF = domG and F(x) = G(x) for all x in domF. Participants explore the implications of the Axiom of Extensionality and the concept of substitutivity of identicals in set theory. The conversation highlights the need for clarity in definitions of functions, referencing "Introduction to Set Theory" by Hrbacek and Jech and "Naive Set Theory" by Paul Halmos as key resources. The conclusion emphasizes the equivalence of function equality to the conditions of domain and output equality.

PREREQUISITES
  • Understanding of set theory concepts, particularly function definitions.
  • Familiarity with the Axiom of Extensionality.
  • Knowledge of domain and range in the context of functions.
  • Experience with logical equivalences and implications in mathematical proofs.
NEXT STEPS
  • Study the Axiom of Extensionality in detail.
  • Learn about function definitions in "Introduction to Set Theory" by Hrbacek and Jech.
  • Explore logical equivalences and implications in mathematical proofs.
  • Review exercises and solutions in "Naive Set Theory" by Paul Halmos.
USEFUL FOR

Mathematics students, educators, and anyone interested in deepening their understanding of set theory and function equality. This discussion is particularly beneficial for those studying formal definitions and proofs in mathematical logic.

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Homework Statement



Prove:

Let F and G be functions. F = G if and only if domF = domG and F(x) = G(x) for all x \in domF.

Homework Equations



?

The Attempt at a Solution



If F = G, then (xFy if and only if xGy) (Substitutivity of identicals)

If (xFy if and only if xGy), then F = G (Axiom of extensionality)

So F = G if and only if (xFy if and only if xGy)

Since F and G are functions,

For all x \in domF, xFy if and only if y = F(x)
For all x \in domG, xGy if and only if y = G(x)

xFy if and only if x \in domF and xFy
xGy if and only if x \in domG and xGy

xFy if and only if x \in domF and y = F(x)
xGy if and only if x \in domG and y = G(x)

So

F = G if and only if ((x \in domF and y = F(x)) if and only if (x \in domG and y = G(x)))

which is equivalent to

F = G if and only if ((x \in domF and x \in domG and y = F(x) and y = G(x)) or (x \in domF and x \in domG and y \neq F(x) and y \neq G(x)) or (x \notin domF and x \notin domG) or (x \in domF and x \notin domG and y \neq F(x)) or (x \notin domF and x \in domG and y \neq G(x))

I was hoping I could show that domF = domG and F(x) = G(x) for all x \in domF was equivalent to this but I couldn't. This is as close as I could get:

(domF = domG and F(x) = G(x) for all x \in domF) if and only if (x \in domF if and only if x \in domG and, if y \in domF, then y = F(x) if and only if y = G(x))

(domF = domG and F(x) = G(x) for all x \in domF) if and only if (((x \in domF and x \in domG and y = F(x) and y = G(x)) or (x \in domF and x \in domG and y \neq F(x) and y \neq G(x)) or (x \notin domF and x \notin domG))

This is close to what I had above but not quite.

I don't know. I think I need to get a set theory book that has solutions; I am currently working through "Introduction to Set Theory" by Hrbacek and Jech. I might switch to "Naive Set Theory" by Paul Hamos; it doesn't even have exercises I don't think.

Any help would be appreciated.
 
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May I ask you how you defined "function"? Is this an exercise of Hrbacek and Jech, if so, which one? I need to see all the definitions before I can answer, since a function is probably something different for me...

On a related note, Hrbacek and Jech is a very good book, I really recommend it. Halmos is more about naive set theory, and is less formal. If you want to learn modern set theory, then Hrbacek and Jech is the best. If you just want to use the set theory somewhere, then I guess Halmos is easier...

Also I would like to add this reference: staff.science.uva.nl/~vervoort/AST/ast.pdf The theory in there is not at all easy, but it has some good and fun exercises (with solutions)
 
Dont mind the above post, I checked Hrbacek and Jech and found it... For the record, I don't really like that definition...

So you indeed have

F=G~\text{iff}~ (xFy ~\Leftrightarrow~xGy)

The domain of F is

dom(F)=\{x~\vert~\exists y:~xFy\}=\{x~\vert~\exists y:~xGy\}=dom(G)

So the domains are equal. Now take an x in dom(F). Then there exists a y such that xFy=xGy. This means exactly that F(x)=G(x) (since y is unique).
 

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